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Physics 111: Mechanics Lecture 2 Wenda Cao NJIT Physics Department.

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Presentation on theme: "Physics 111: Mechanics Lecture 2 Wenda Cao NJIT Physics Department."— Presentation transcript:

1 Physics 111: Mechanics Lecture 2 Wenda Cao NJIT Physics Department

2 September 8, 2008 Motion along a straight line  Motion  Position and displacement  Average velocity and average speed  Instantaneous velocity and speed  Acceleration  Constant acceleration: A special case  Free fall acceleration

3 September 8, 2008 Motion  Everything moves!  Motion is one of the main topics in Physics 111  Simplification: Moving object is a particle or moves like a particle: “point object”  Simplest case: Motion along straight line, 1 dimension LAX Newark

4 September 8, 2008 One Dimensional Position x  What is motion? Change of position over time.  How can we represent position along a straight line?  Position definition: Defines a starting point: origin (x = 0), x relative to origin Direction: positive (right or up), negative (left or down) It depends on time: t = 0 (start clock), x(t=0) does not have to be zero.  Position has units of [Length]: meters. x = m x = - 3 m

5 September 8, 2008 Vector and Scalar  A vector quantity is characterized by having both a magnitude and a direction. Displacement, Velocity, Acceleration, Force … Denoted in boldface type with an arrow over the top.  A scalar quantity has magnitude, but no direction. Distance, Mass, Temperature, Time …  For the motion along a straight line, the direction is represented simply by + and – signs. + sign: Right or Up. - sign: Left or Down.  2-D and 3-D motions.

6 September 8, 2008 Quantities in Motion  Any motion involves three concepts Displacement Velocity Acceleration  These concepts can be used to study objects in motion.

7 September 8, 2008 Displacement  Displacement is a change of position in time.  Displacement: f stands for final and i stands for initial.  It is a vector quantity.  It has both magnitude and direction: + or - sign  It has units of [length]: meters. x 1 (t 1 ) = m x 2 (t 2 ) = m Δx = -2.0 m m = -4.5 m x 1 (t 1 ) = m x 2 (t 2 ) = m Δx = +1.0 m m = +4.0 m

8 September 8, 2008 Distance and Position-time graph  Displacement in space From A to B: Δx = x B – x A = 52 m – 30 m = 22 m From A to C: Δx = x c – x A = 38 m – 30 m = 8 m  Distance is the length of a path followed by a particle from A to B: d = |x B – x A | = |52 m – 30 m| = 22 m from A to C: d = |x B – x A |+ |x C – x B | = 22 m + |38 m – 52 m| = 36 m  Displacement is not Distance.

9 September 8, 2008 Velocity  Velocity is the rate of change of position.  Velocity is a vector quantity.  Velocity has both magnitude and direction.  Velocity has a unit of [length/time]: meter/second.  Definition: Average velocity Average speed Instantaneous velocity

10 September 8, 2008 Average Velocity  Average velocity  It is slope of line segment.  Dimension: [length/time].  SI unit: m/s.  It is a vector.  Displacement sets its sign.

11 September 8, 2008 Average Speed  Average speed  Dimension: [length/time], m/s.  Scalar: No direction involved.  Not necessarily close to V avg : S avg = (6m + 6m)/(3s+3s) = 2 m/s V avg = (0 m)/(3s+3s) = 0 m/s

12 September 8, 2008 Graphical Interpretation of Velocity  Velocity can be determined from a position-time graph  Average velocity equals the slope of the line joining the initial and final positions. It is a vector quantity.  An object moving with a constant velocity will have a graph that is a straight line.

13 September 8, 2008 Instantaneous Velocity  Instantaneous means “at some given instant”. The instantaneous velocity indicates what is happening at every point of time.  Limiting process: Chords approach the tangent as Δt => 0 Slope measure rate of change of position  Instantaneous velocity:  It is a vector quantity.  Dimension: [Length/time], m/s.  It is the slope of the tangent line to x(t).  Instantaneous velocity v(t) is a function of time.

14 September 8, 2008  Uniform velocity is constant velocity  The instantaneous velocities are always the same, all the instantaneous velocities will also equal the average velocity  Begin with then Uniform Velocity x x(t) t0 xixi xfxf v v(t) t0 tftf vxvx titi

15 September 8, 2008 Average Acceleration  Changing velocity (non-uniform) means an acceleration is present.  Acceleration is the rate of change of velocity.  Acceleration is a vector quantity.  Acceleration has both magnitude and direction.  Acceleration has a unit of [length/time 2 ]: m/s 2.  Definition: Average acceleration Instantaneous acceleration

16 September 8, 2008 Average Acceleration  Average acceleration  Velocity as a function of time  When the sign of the velocity and the acceleration are the same (either positive or negative), then the speed is increasing  When the sign of the velocity and the acceleration are in the opposite directions, the speed is decreasing  Average acceleration is the slope of the line connecting the initial and final velocities on a velocity-time graph

17 September 8, 2008 Instantaneous and Uniform Acceleration  The limit of the average acceleration as the time interval goes to zero  When the instantaneous accelerations are always the same, the acceleration will be uniform. The instantaneous acceleration will be equal to the average acceleration  Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph

18 September 8, 2008 Relationship between Acceleration and Velocity  Velocity and acceleration are in the same direction  Acceleration is uniform (blue arrows maintain the same length)  Velocity is increasing (red arrows are getting longer)  Positive velocity and positive acceleration

19 September 8, 2008 Relationship between Acceleration and Velocity  Uniform velocity (shown by red arrows maintaining the same size)  Acceleration equals zero

20 September 8, 2008 Relationship between Acceleration and Velocity  Acceleration and velocity are in opposite directions  Acceleration is uniform (blue arrows maintain the same length)  Velocity is decreasing (red arrows are getting shorter)  Velocity is positive and acceleration is negative

21 September 8, 2008 Kinematic Variables: x, v, a  Position is a function of time:  Velocity is the rate of change of position.  Acceleration is the rate of change of velocity.  Position Velocity Acceleration  Graphical relationship between x, v, and a An elevator is initially stationary, then moves upward, and then stops. Plot v and a as a function of time.

22 September 8, 2008 Motion with a Uniform Acceleration  Acceleration is a constant  Kinematic Equations

23 September 8, 2008 Notes on the Equations  Given initial conditions: a(t) = constant = a, v(t=0) = v 0, x(t=0) = x 0  Start with  We have  Shows velocity as a function of acceleration and time  Use when you don’t know and aren’t asked to find the displacement

24 September 8, 2008  Given initial conditions: a(t) = constant = a, v(t=0) = v 0, x(t=0) = x 0  Start with  Since velocity change at a constant rate, we have  Gives displacement as a function of velocity and time  Use when you don’t know and aren’t asked for the acceleration Notes on the Equations

25 September 8, 2008  Given initial conditions: a(t) = constant = a, v(t=0) = v 0, x(t=0) = x 0  Start with  We have  Gives displacement as a function of time, initial velocity and acceleration  Use when you don’t know and aren’t asked to find the final velocity Notes on the Equations

26 September 8, 2008  Given initial conditions: a(t) = constant = a, v(t=0) = v 0, x(t=0) = x 0  Start with  We have  Gives velocity as a function of acceleration and displacement  Use when you don’t know and aren’t asked for the time Notes on the Equations

27 September 8, 2008 Problem-Solving Hints  Read the problem  Draw a diagram Choose a coordinate system, label initial and final points, indicate a positive direction for velocities and accelerations  Label all quantities, be sure all the units are consistent Convert if necessary  Choose the appropriate kinematic equation  Solve for the unknowns You may have to solve two equations for two unknowns  Check your results Estimate and compare Check units

28 September 8, 2008 Free Fall Acceleration  Earth gravity provides a constant acceleration. Most important case of constant acceleration.  Free-fall acceleration is independent of mass.  Magnitude: |a| = g = 9.8 m/s 2  Direction: always downward, so a g is negative if define “up” as positive, a = -g = -9.8 m/s 2  Try to pick origin so that x i = 0 y

29 September 8, 2008 Free Fall Acceleration  Two important equation:  Begin with t 0 = 0, v 0 = 0, x 0 = 0  So, t 2 = 2|x|/g same for two balls!  Assuming the leaning tower of Pisa is 150 ft high, neglecting air resistance, t = (2  150  0.305/9.8) 1/2 = 3.05 s x 0

30 September 8, 2008 Summary  This is the simplest type of motion  It lays the groundwork for more complex motion  Kinematic variables in one dimension Positionx(t)mL Velocityv(t)m/sL/T Accelerationa(t)m/s 2 L/T 2 All depend on time All are vectors: magnitude and direction vector:  Equations for motion with constant acceleration: missing quantities x – x 0 v t a v 0


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